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Math Help - Help with Proof of Sets

  1. #1
    Junior Member
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    Help with Proof of Sets

    Prove  (A \cup B) - (A \cap B) = (A-B) \cup (B-A)

    (A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)
    Let  x \in (A \cup B) - (A \cap B). Then x \in A \cup B and  x \notin A \cap B. Then x \in A or x \in B but x \notin A and x \notin B. Since x \in A and x \notin B then  x \in (A-B). Also since x \in B and x \notin A then, x \in (B-A). Now  x \in (A-B)\cup(B-A).

    Therefore, (A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)

    Do I have the general idea right? What can I fix? I'm sure I did something wrong. Yes, I also realized I skipped the other part of the proof.
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  2. #2
    Super Member

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    Hello, JSB1917!

    I used a two-column proof . . .


    \text{Prove: }\;(A \cup B) - (A \cap B) = (A-B) \cup (B-A)

    \begin{array}{ccccccc}<br />
1. & (A \cup B) - (A \cap B) && 1.& \text{ Given} \\ \\<br />
2. & (A \cup B) \cap (\overline{A \cap B}) && 2. & \text{d{e}f. Subtraction} \\ \\<br />
3. & (A \cup B) \cap (\overline A \cup \overline B) && 3. & \text{DeMorgan's Law} \\ \\<br />
4. & (A \cap \overline A) \cup (A \cap \overline B) \cup (B \cap \overline A) \cup (B \cap \overline B) && 4. & \text{Distributive Prop.} \\ \\<br />
5. & \emptyset \;\;\cup (A \cap \overline B) \cup (B \cap \overline A) \cup \;\;\emptyset && 5. & S \cap S \:=\:\emptyset \\ \\<br />
6. & (A \cap \overline B) \cup (B \cap \overline A) && 6. & S \cup \emptyset \:=\:S \\ \\<br />
7. & (a - B) \cup (B - A) && 7. & \text{d{e}f. Subtraction}<br />
\end{array}
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